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Metamath Proof Explorer

Theorem islmim2

Description: An isomorphism of left modules is a homomorphism whose converse is a homomorphism. (Contributed by Mario Carneiro, 6-May-2015)

Ref Expression
Assertion islmim2 
|- ( F e. ( R LMIso S ) <-> ( F e. ( R LMHom S ) /\ `' F e. ( S LMHom R ) ) )

Proof

Step Hyp Ref Expression
1 eqid 
 |-  ( Base ` R ) = ( Base ` R )
2 eqid 
 |-  ( Base ` S ) = ( Base ` S )
3 1 2 islmim 
 |-  ( F e. ( R LMIso S ) <-> ( F e. ( R LMHom S ) /\ F : ( Base ` R ) -1-1-onto-> ( Base ` S ) ) )
4 1 2 lmhmf1o 
 |-  ( F e. ( R LMHom S ) -> ( F : ( Base ` R ) -1-1-onto-> ( Base ` S ) <-> `' F e. ( S LMHom R ) ) )
5 4 pm5.32i 
 |-  ( ( F e. ( R LMHom S ) /\ F : ( Base ` R ) -1-1-onto-> ( Base ` S ) ) <-> ( F e. ( R LMHom S ) /\ `' F e. ( S LMHom R ) ) )
6 3 5 bitri 
 |-  ( F e. ( R LMIso S ) <-> ( F e. ( R LMHom S ) /\ `' F e. ( S LMHom R ) ) )